divisibility rules — a note by arron kau _ brilliant

5/20/2018 Divisibility Rules a Note by Arron Kau _ Brilliant

1

7/12/2014 Divisibility Rules a note by Arron Kau | Brilliant

https://brilliant.org/discussions/thread/divisibility-rules/

When learning about multiples and divisors, there are several rules of divisibil ity that a student may

encounter. Below, we list some famous rules of divisibi lity:

When dividing by...

2:The last digit is even.

3:The sum of digits is a multiple of 3.

4:The last 2 numbers are a multiple of 4.

5:The last digit is either 0 or 5.

6:Multiple of 2 and 3

8:The last 3 digits are a multi ple of 8.

9:The sum of digits is a multiple of 9.

10:The last digit is 0.

11:The alternating sum of digits is a multiple of 11.

We will show the proofsof rules of 8 and 11. The rest follow in a similar manner.

Proof: When a number is divisible by 8, the last 3 digits are a multiple of 8.

Let the number be , where are digits and is a non-

negative integer.

Clearly, is a multiple of 8, since . Hence, is a multiple of 8 if and only if

is a multiple of 8.

Proof: When a number is divisible by 11, the alternating sum of digits is a multiple of

11.From Factorization , we know that

is always a multiple of 11.

Hence, if , then

Since the terms in the square brackets consist of multiples of 11, it follows that is a

multiple of 11 if and only if the alternating sum is a multiple of 11.

Divisibility RulesShared by Arron Kau32, USA 3 months, 2 weeks ago

Use Brilliant to build your problem

solving skills in mathand science

Sign up! or Log in
https://brilliant.org/https://brilliant.org/account/login/?next/discussions/thread/divisibility-rules/https://brilliant.org/account/signup/?next=/discussions/thread/divisibility-rules/https://brilliant.org/https://brilliant.org/profile/arron-udsft3/feed/


2



About Help Terms Privacy Brilliant 2014

Sort by:

#NumberTheory #RulesOfDivisibility #KeyTechniques

APPLICATION AND EXTENTIONS

Find all possible values of such that the number is a multiple of 3.

From the rules of divisibility, the number is a multiple of 3 if and only if the sum of

the digits is a multiple of 3. Since , this implies that

are all the possible values.

Show that if the last 3 digits of a number are , then is a multiple of 8 if and

only if is a multiple of 8.

This follows because . Hence, by the divisibility

rule of 8, is a multiple of 8 if and only if is a multiple of 8 if and only if is

a multiple of 8.

Divisibility rule of 7: Break up the number into blocks of 6, starting from the right. Add

up all these blocks, and the resultant number has to be a multiple of 7.

This follows because , so

is a multiple of 7 if and only if

is a multiple of 7. If we use , we get the result as

stated.

Show that the 6 digit number is a multiple of 7 if and only if

is a multiple of 7.

Solution: This follows because

Hence is a multiple of 7 if and only if is a multiple of 7.

newest

Anuj Shikarkhane

Thanks for this

Jun 26

top
https://brilliant.org/discussions/thread/divisibility-rules/?sort=tophttps://brilliant.org/profile/anuj-84q655/feed/https://brilliant.org/discussions/thread/divisibility-rules/?sort=newhttps://brilliant.org/tags/keytechniques/https://brilliant.org/tags/rulesofdivisibility/https://brilliant.org/tags/numbertheory/https://brilliant.org/privacy_policy/https://brilliant.org/terms-of-use/https://brilliant.org/help/https://brilliant.org/about/


3


https://play.google.com/store/apps/details?id=org.brilliant.android&referrer=utm_source%3Dbrilliant%26utm_medium%3Dsite%26utm_campaign%3Dplaystore_full_footer_icon

divisibility rules — a note by arron kau _ brilliant

Documents